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Problem

Power-log

Consider the family of functions

$ x^p \ln(x) $.

Part 1: Calculate the indefinite integral

$ \int x^p \ln(x) dx $.

Part 2: For what values $ p $ does the following integral converge?

$ \int_0^1 x^p \ln(x) dx $

Solution

Part 1

This family of functions is the product of a power function with the natural logarithm function. Therefore the method of integration by parts should be used.

According to LIATE, let $ u = \ln(x) $ and $ dv = x^p dx $.

Then $ du = \frac{1}{x}dx $ and $ \frac{{x}^{p+1}}{p+1} $.

Therefore,

$ \int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p+1}}{p+1} \frac{1}{x} dx $
$ \int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p}}{p+1} dx $
$ \int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \frac{{x}^{p+1}}{(p+1)^2} + C $

Part 2

From the result of the indefinite integral, it is obvious that $ p \ne -1 $. Just look at the denominators. Now this means

$ \int_0^1 x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} \Biggr|_0^1 - \frac{{x}^{p+1}}{(p+1)^2} \Biggr|_0^1 $.

Notice that if $ p < -1 $, the power functions become reciprocal functions, which means the integral diverges when evaluating the lower bound. However if $ p > -1 $, the integral converges to the expression

$ \int_0^1 x^p \ln(x) dx = \frac{-1}{(p+1)^2} $.

Therefore the integral

$ \int_0^1 x^p \ln(x) dx $

converges if $ p > -1 $.

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